Being keen in Esperanto I enquired into the opportunity to participate in an exchange program with someone from Esperantoland, because I thought it would an enriching experience.
Performing the ##x## integration exactly for ##Z[w]## could very well not be possible. You may very well not be able to obtain the exact value even if you were just considering this integral:
\begin{align*}
\frac{1}{(2 \pi)^{n/2}} \int d^nx \exp \left( - \frac{1}{2} w_i (x) w_i (x) \right)...
@DrClaude you do a shift in the summation variable ##n## followed by extending the sum:
##\sum_{n=1}^\infty n \frac{1}{2 (n+1)^2} = \sum_{n=2}^\infty (n-1) \frac{1}{2 n^2} = \sum_{n=1}^\infty (n-1) \frac{1}{2 n^2}##.
No, your idea worked I think.
There are no ##u## such that ##H u \not=0## and ##H^\dagger H u =0##, so ##\ker (H) = \ker(H^\dagger H)##. It follows then from ##{H^\dagger}^2 = H^\dagger H## that ##\ker (H) = \ker ({H^\dagger}^2)##.
The second question is are there ##u## such that ##H^\dagger u...
It would be good there was an easy way of proving ##\ker (H) = \ker (H^\dagger)## as this was the bulk of my solution in post #2.
If I understand you you are saying if ##\dim \ker (H) = \dim \ker (H^\dagger)## then does ##{H^\dagger}^2 \ker (H) = H^\dagger H \ker (H)## imply ##\ker (H) = \ker...
I get:
\begin{align*}
\| Hx -H^\dagger x \|^2 = \langle (H H^\dagger - H^\dagger H) x , x \rangle
\end{align*}
Do we immediately know ##H## is a normal matrix?
The sum of residues is indeed equal to:
\begin{align*}
\dfrac{\sin \left( \frac{3 \pi}{8} \right)}{\sin \left( \frac{3 \pi}{4} \right)}
\end{align*}
You then just substitute ##\sin \left( \frac{3 \pi}{4} \right) = \frac{1}{\sqrt{2}}## into it to obtain the desired result.
The denominator of the 2nd term is the negative of the denominator of the first term. That allows you to write your expression as a single fraction.
And then, yes, you use ##\sin (x) = \frac{e^{i x} - e^{-i x}}{2i}##.
Yes, you get out the additional result:
\begin{align*}
\int_0^\infty \frac{1-x^2}{1+x^4} dx = 0
\end{align*}
You can see this also by doing ##y=1/x## in the following integral:
\begin{align*}
\int_0^\infty \frac{1}{1+x^4} dx = \int_0^\infty \frac{y^2}{1+y^4} dy
\end{align*}
Because you have a second order differential equation the general solution will involve two unknown constants, so you need ##x(0)## and ##x'(0)##. Now, ##x'(0)## was not specified in the question but that does not present a problem because you are given a formula for ##x'(t)##. Yep?
Do you know...
In the previous post I wrote down slightly the wrong matrix. I've now written down the correct matrix and recalculated the eigenvalues.
By the way, there are other methods of solving the problem!
I am not contributing something new. I am taking a result given in Watson and arriving at the same condition that @renormalize arrived at.
On page 148 of Watson is the formula:
\begin{align*}
J_\mu (az) J_\nu (bz) & = \dfrac{ (\frac{1}{2} az)^\mu (\frac{1}{2} b z)^\nu}{\Gamma (\nu+1)}...
It's not exactly "drivng me batty", but let's switch from ##e## to ##x## anyway. Say we have
\begin{align*}
A J_2 (x) + B Y_2 (x)= \frac{1}{x} \sum_{n=1}^\infty \dfrac{J_n (n x)}{n} \dfrac{J_{n+1} ((n+1) x)}{n+1}
\end{align*}
We have that ##J_2 (0) = 0## and the RHS is zero at ##x=0## (because...
I know you are considering ##0 < e < 0##. But can we not use that ##\lim_{e \rightarrow 0} J_2 (e) \rightarrow 0## and that ##Y_2 (e)## is singular for ##e=0##?
Wouldn't the sum:
\begin{align*}
\frac{1}{e} \sum_{i=1}^\infty \dfrac{J_i ( i e)}{i} \dfrac{J_{i+1} ((i+1) e)}{i+1}
\end{align*}...
I've just remembered, there's this BBC children's programme called Blue Peter which featured the "science sisters" for a while. I was actually doing my PhD with one of the science sisters at the time! It was so surreal to see her on TV. Blue Peter encouraged you to write in with questions. I...
John Baez has been posting here from 2004 to 2021:
https://www.physicsforums.com/search/7497600/
unless that is somebody pretending to be John Baez.
Francesca Vidotto, who wrote the book "Covariant loop quantum gravity" with Carlo Rovelli...
Your matrix pertains to 2-spinors, and not Dirac 4-spinors. If you actually want to learn about these 2-spinors, which have applications outside of QM, they have applications in general relativity, you can look at chapter 41 of GRAVITATION Misner et al or you could look at chapter 2 of Advanced...